Q. A truck moving with a speed of 36 km/hr is stopped by applying breaks after traveling 2 m. If the same truck moves with a speed of 72 km/hr, then find its stopping distance (Assume constant retardation).
Answers
Correct option is
D
8m
Stopping distance s=
2a
u
2
s
1
s
2
=
u
1
2
u
2
2
⇒s
2
=[
u
1
u
2
]
2
s
1
⇒s
2
=(
40
80
)
2
2⇒s
2
=8m
The stopping distance of the truck is 8 m.
Given: A truck moving with a speed of 36 km/hr is stopped by applying breaks after traveling 2 m.
To Find: The stopping distance of the truck if it moves with a speed of 72 km/hr.
Solution:
We can solve the numerical by making use of the equations of motions which states that,
v² = u² + 2aS ...(1)
Where v = final velocity, u = initial velocity, a = acceleration, S = Distance.
Coming to the numerical,
It is said that a truck moving with a speed of 36 km/hr is stopped by applying breaks after traveling 2 m. So, we can imply that;
The final velocity (v) = 0
The initial velocity (u) = 36 km/hr
The distance traveled = 2 m = 0.002 km
So, putting the respective values in (1), we shall find the acceleration;
v² = u² + 2aS
⇒ 0 = 36² + 2 × a × 0.002
⇒ - 1296 = a × 0.004
⇒ a = - 1296 / 0.004
⇒ a = - 324000 km/hr²
Since the acceleration is negative, so there is retardation.
For the second case, we are told to assume the retardation to be constant, so we can say that;
The final velocity (v) = 0
The initial velocity (u) = 72 km/hr
The retardation (a) = - 324000 km/hr²
So, putting the respective values in (1), we shall find the stopping distance;
v² = u² + 2aS
⇒ 0 = 72² + 2 × ( - 324000 ) × S
⇒ - 5184 = - 648000 × S
⇒ S = 5184 / 648000
⇒ S = 8 × 10^-3 km
= 8 m
Hence, the stopping distance of the truck is 8 m.
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